Why does $\bar{t_n}-{T_n}=\frac{3}{4^{k_n}}$

Question

We have that $$t=0_4.q_1q_2q_3... \in [0,1]$$ t is in Quaternary form.

Let $$T_n=0_4.q_1q_2q_3... q_{k_{n-1}}0$$ and

$$\bar{t_n}=0_4.q_1q_2q_3..q_{k_{n-1}}3$$

Why does $\bar{t_n}-{T_n}=\frac{3}{4^{k_n}}$?

Answer 1

Suppose $a\in[0,1)$ has the following form in base $b$ $$ a=0.a_1a_2\cdots a_n $$ The above notation is just short-hand for $$ a=\sum_{j=1}^na_jb^{-j} $$ So in your case $$ T_n=\sum_{j=1}^{k_{n-1}}{q_j\over4^j},\quad \bar t_n=\sum_{j=1}^{k_{n-1}}{q_j\over4^j}+{3\over4^{k_{n-1}+1}} $$ If $k_n=k_{n-1}+1$, the above equations immediately yield the desired result.